Just like you can remember a list of items in sequence by storing them at a sequence of places (eg.The BLOKES System places), you can memorise a pack of cards by having a visual image that means each card of a pack; and you might even imagine a jester to mean the joker card.
One way to do this is to look at the 00 to 99 people images and pick 52 people from there.
Below is a list of consonants and vowels which can construct syllables to match syllables in the 00 to 99 People System; so the 3 of Clubs can be consonant 3 and the vowel for Clubs:
3 is G (see below; Clubs is O. So look up GO person in the 00 to 99 People System: GOrdon.
| Prefix | Card |
| Ch | Ace |
| Sh | 2 |
| G | 3 |
| D | 4 |
| N | 5 |
| S | 6 |
| Z | 7 |
| T | 8 |
| F | 9 |
| P | 10 |
| J | Jack |
| L | Queen [Lady] |
| K | King |
| Spades | A |
| Hearts | E |
| Diamonds | I |
| Clubs | O |
Note that I chose consonants that are easy to differentiate phonetically from each other; So M is not used since N is used.
When I implement a system of over 2500 cards (see the article about a 5000 system), I will be on my way to representing two cards in one image. It's unnecessary but would make memorisation faster. In memory sports, people want to shave time off how long it takes them to memorise a pack of cards. So using one image to mean two cards does / would benefit those competitors. Although SH, S, Z, T, P and N syllables are in the table above, they are not expressed in the 5000 letter pair system article as left side syllables; so you would not find SATI or PAZO , for instance, in the letter pairs of the 5000 person 5K system article. Note that the 5000 people system gives a two syllable name to 5000 people; and if each card is treated as a syllable then a pair of cards can match up with one person whose name uses those syllables. Maybe I would use people from the Shakepeare article or animals from the Animal people system to substitute in for those missing sets of 52.
A question which jumps out is: if I am using a peg location image as the place where a card's visual image occurs, can I ever re-use the location for future memorising of something else?
Yes, the aim is for natural forgetfulness to make you very quickly forget the story where the item that represents, say, the Ace of Hearts, occurs at a location like a garden shed. And that implies that, using memory techniques to memorise exam prompts is not a one time exercise: there needs to be revisiting of the imagined scenes to make your memory stronger of which visuals occur at which locations. After the exam, if you stop thinking about that visual story then the story should be gradually forgotten naturally.
Another question is about the picturing of more than one visual item at the same location. You can do that but bear in mind that there is a simplicity to just storing one visdual prompt at one location; and bear in mind that, once a lot of visual images are imagined at one location, it is more of a mental strain. But I definitely think it is reasonable to imagine about 3 exam prompts at a single scene location.
Memorising Cards for Magic Tricks
The letter pairs article presents the idea that one image can represent two letters of the alphabet from AA , AB, AC, ... through to ZZ. There are 26 types of 52 playing cards (I explain below); and so the same system can be used to memorise cards:
If you think of Black Aces as the letter A, Black 7s as the letter G; then the image for AG could also be representing 'a black Ace followed by a black 7'. This would be more rapid than memorising cards one by one but also less specific because, for instance, the recalled Black Ace might be Clubs or Spades - you don't know.
But you could pre-prepare 26 cards of the pack where there is only one black Ace, only one black 7, etc.; and then, if you memorise from that pile of cards, you are left with no doubt which black Ace you are recalling and which black 7 you are recalling.
So you could look, to the spectator, like you are counting out 26 cards but you are really memorising them rapidly; and that leads on to the trick you do: you know every card in sequence in that pile of 26 cards.
Here is a video about it. I made the video before I finalised the following list of card piles:
(so do not expect the cards in the video to match the newer list below)
| Ch | Black | Ace |
| Sh | Black | 2 |
| G | Black | 3 |
| D | Black | 4 |
| N | Black | 5 |
| S | Black | 6 |
| Z | Black | 7 |
| T | Black | 8 |
| F | Black | 9 |
| P | Black | 10 |
| J | Black | Jack |
| L | Black | Queen |
| K | Black | King |
| A | Red | Ace |
| M | Red | 2 |
| C | Red | 3 |
| U | Red | 4 |
| E | Red | 5 |
| Y | Red | 6 |
| V | Red | 7 |
| H | Red | 8 |
| I | Red | 9 |
| O | Red | 10 |
| W | Red | Jack |
| Q | Red | Queen |
| R | Red | King |
Three times Three times Three is 27
This section is about how the letter pairs article is a way to memorise 3 cards per letter so that a letter pair represents 3+3=6 cards. I do not mean that you learn a card in terms of suit and numeral but that you learn a choice of 3 options about each card. "Eg. Is it an Ace, a royal card or a numeral?"
A letter pair image can represent 6 cards if you have not just A to Z but a 27th option as well: 27 x 27. Eg. If Sh is thought of as a letter; and you could have ShA, ShB, ShC, .... ShX, ShY, ShZ, ShSh as letter pairs.
Well, not specific cards but three choices of card such as 'Higher, Lower, the same' or 'Numeral, Royal card, Ace'. You could represent any 3 combinations as 27 choices (one of 27 letters); so a letter pair would b3 three choices plus three choices.
So you could memorise 18 cards in terms of whether they are each higher / lower / the same as the previous card; and that 18 sequence would be just 6 + 6 + 6 choices; and so it would be represented visually by three 'letter pair' images.
More three choices:
Red / Black / Joker
Ace of Hearts / Ace of Diamonds / Other card
| Three Lots of Three Choices | |
| O | 111 |
| I | 112 |
| K | 113 |
| A | 121 |
| E | 122 |
| F | 123 |
| G | 131 |
| H | 132 |
| B | 133 |
| M | 211 |
| N | 212 |
| L | 213 |
| S | 221 |
| C | 222 |
| D | 223 |
| P | 231 |
| Q | 232 |
| R | 233 |
| J | 311 |
| T | 312 |
| U | 313 |
| V | 321 |
| W | 322 |
| X | 323 |
| Y | 331 |
| Z | 332 |
| SH | 333 |
Spotting the Aces
Using the three options of 'Black Ace, Red Ace, Other card', you could quickly look through a pack of cards and memorise (using 6 facts per letter pair) which cards are aces.
Another way of mentally marking the Aces is to have 4 actions. As you look through the pack, if you see an Ace then you can imagine the person who represents that ordinal position in the pack and imagine an action that represents either Aces of Spades / Ace of Hearts / Ace of Diamonds / Ace of Clubs.
So, if at card 33, there is an Ace of Hearts, you would imagine the Ace of Hearts action happening to the standard person image that you use to represent the number 33.
The Spades action could be a person doing a digging motion with a spade.
The Hearts action could be a Valentine's heart throbbing at the person's chest.
The Diamonds action could be a person throwing diamonds into the air joyfullly.
The Clubs action could be a person swinging a baseball bat with a whoosh sound.
Card Counting to Deduce the card removed from a 52 card pack
If you remove a card from a pack and then look through the cards, I think you can do a logical technique to deduce the removed card:
If each card has a different number value from 1 to 52 then you would expect the grand total to be a constant answer. Whatever card is removed from the pack has a unique effect on the sum of the 51 remaining cards. That is the principle on which a maths can be used to deduce a missing card from a pack.
A letter pair can store 5 x 5 x 5 x 5 options
You already saw how a card could represent 3 x 3 x 3 options such as 'Higher / Lower / Same' or 'Royal / Ace / Numeral'. A letter between A and Y can represent 5x5 choices (=25 options [so 25 letters]). A letter pair is then able to hold 5, 5, 5 and 5 options because it is two letters rather than the 5x5 limit of just one letter.
A subset of that system would be the ability to memorise 4x4x4x4 options by using a letter pair. That creates an interesting possibility involving the two halves of the pack that I listed earlier: where each pile of 26 only has one black 7, not two black 7s, etc.. That is 2 options: does a card which you pick up belong in pile 1 or in pile 2? But you can have a further question: is the card red or black? In that way, a card which you pick up is one of four options: Pile 1 Black, Pile 1 Red, Pile 2 Black, Pile 2 Red. If you memorise the cards which you pick up and memorise which of the 4 options each card is then you could do some good tricks. You could deal out the gathered cards into four seemingly random piles: really piles of Pile 1 Black, Pile 1 Red, Pile 2 Black, Pile 2 Red; and, from there, do a trick that depends on knowing the colour of a card or demonstrate all the reds together and all the blacks together. After that, two of the 4 piles can come together as pile 1 and be shuffled; and two of the 4 piles can be gathered together into pile 2 and be shuffled. Since you know the cards of each pile, you can do tricks where a card from the other pile is introduced to the other pile; and you find it easy to point out which card has been added.
The 21 example
I am trying to explain the 5 x 5 x 5 x 5 idea more clearly. A letter pair would be something like AA or YY or AY or BC or GN. This course offers ways to represent each letter pair as a single image of maybe a person or a household item. Once you know a system of images like that, you can memorise letter pairs easily.
As you look through a pack of cards, each letter pair represents 5 x 5 x 5 x 5 infromation: for every four cards which you look at, you are memorising one of five states about each of the four cards.
In a 'game of 21' card trick, the five states can be:
The card is a 10 or Jack or Queen or King
The card is an Ace.
The card is a 2.
The card is a 9.
The card is any other card.
So you could memorise 36 cards at 9 peg locations where each location holds a letter pair image. And that means four cards' information: any letter on its own represents two cards in terms of those 5 states; and a pair of letters means 4 such states; and one letter pair image is 4 states; and 9 of those is 36 catds' states. The 36 cards can be dealt out into 3 piles; and then you can recall the cards three at a time and state if the cards will be 21 or will not be 'bust' or will be 'bust'. You might even say that the first two cards add up to 21, then reveal them [an Ace and a King perhaps], then say that the third card is still 21; and turn over a Jack.
With the remaining 16 unused cards of the pack, you could ask the person to arrange them in sequence from lowest through to highest - where Ace is highest. If you have a way to memorise the count of each of the 36 cards revealed then you can deduce the 16 cards.
The animal system article has 10 animal images per type of animal. You can learn five of those animals for thirteen animal types. Eg. Animals A0 to A4 can represent "No Ace revealed yet", "1 Ace revealed while turning over the '21 game' cards", "2 Aces revealed while turning over the '21 game' cards", "3 Aces revealed while turning over the '21 game' cards", and "4 Aces revealed while turning over the '21 game' cards". If your short term memory is good then, after revealing the 36 cards, you will know what is the current version of each of the 13 animal types. Maybe the animal type representing Jacks has got as high as J3. So you know that a Jack is left in the unseen pile of 16 cards; and because you know that they are in ascending sequence, you can recall the status of the 2s, then the status of the 3s, the 4s, etc. and announce the cards one by one before turning them over. I thought of the 21 game trick and this 16 card deduction technique in late February 2024. I am really pleased with it.
Note: It is a shame though that, for example, a 7 and a 3 and a 2 would be a mystery: they would all be in the category "The card is any other card."
In a situation where you can not work out "bust", "not bust" and "21", you can ask the audience to make a suggestion what they think the answer is: side-step the difficulty.
Making the card memorisation less obvious
Using a system where a letter pair represents two cards (see early in this article where black cards have 13 letters and red cards have 13 letters), a letter pair does not tell me the identity of 2 cards. Eg. "Red Jack and Black Three" could be Jack of Hearts and 3 of Clubs but it might be Jack of Diamonds and 3 of Spades, etc.
But you can make card tricks where that knowledge is sufficient to do something impressive. Eg. "It's a red card... the Jack of [turn card over as you say it] Hearts!
That looks like you knew it even though you only know the suit as you turn the card over.
That type of letter pair could be used in the following example:
You say that you need to make sure that there are 52 cards; and you look through them, memorising the Aces and their positions in the pack. The audience just think that you were counting the cards. When you count through the middle of the pack, you can memorise 4 letter pairs that represent 8 cards such as "Red Jack" or "Black Three". This gives you speed and it looks less like you are memorising anything.
With the pack turned over, you can cut the pack and be quite accurate at it with practice. If you made a minor error in the cut then the card you see as you do the cut will differ; and, because of your mid pack memorisation, you know the number position of that card - at least before the cut happened.
You can then deal the cards out in 4 piles and do a trick where you take out four cards from that display and reveal them all to be Aces.
That involves memorising the whereabouts of each card after a cut of the pack and after dealing into 4 piles - so a bit of research is needed; but it's impressive.
You would learn the whereabouts for 8 scenarios - depending on which of the mid pack 8 cards you saw when you approximately cut the pack in the middle.
There is a risk that two cards of those eight cards are red Jacks - so the trick won't work every time but it will tend to work. Maybe turn it into a joke trick of some kind if it goes wrong.